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\(x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-x^2z^2\left(z-x\right)\)
\(=x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-x^2z^2\left[\left(z-y\right)+\left(y-x\right)\right]\)
\(=x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-x^2z^2\left(z-y\right)-x^2z^2\left(y-x\right)\)
\(=\left(y-x\right)\left(x^2y^2-x^2z^2\right)+\left(z-y\right)\left(y^2z^2-x^2z^2\right)\)
\(=x^2\left(y-x\right)\left(y-z\right)\left(y+z\right)+z^2\left(z-y\right)\left(y-x\right)\left(y+x\right)\)
\(=\left(y-x\right)\left(z-y\right)\left(-x^2y-x^2z+z^2y+z^2x\right)\)
\(=\left(y-x\right)\left(z-y\right)\left[xz\left(z-x\right)+y\left(z-x\right)\left(z+x\right)\right]\)
\(=\left(y-x\right)\left(z-y\right)\left(z-x\right)\left(xy+yz+xz\right)\)
\(x\left(y^2-z^2\right)+y\left(z^2-x^2\right)+z\left(x^2-y^2\right)\)
\(=x\left(y^2-z^2\right)-y\left(y^2-z^2+x^2-y^2\right)+z\left(x^2-y^2\right)\)
\(=\left(y^2-z^2\right)\left(x-y\right)+\left(x^2-y^2\right)\left(z-y\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(y+z-x-y\right)=\left(x-y\right)\left(y-z\right)\left(z-x\right)\)
chúc bn hc tốt ^^
Ta có: \(x\left(y^2-z^2\right)+y\left(z^2-x^2\right)+z\left(x^2-y^2\right)\)
\(=x\left(y-z\right)\left(y+z\right)+yz^2-x^2y+zx^2-y^2z\)
\(=x\left(y-z\right)\left(y+z\right)-\left(y^2z-yz^2\right)-\left(x^2y-zx^2\right)\)
\(=x\left(y-z\right)\left(y+z\right)-yz\left(y-z\right)-x^2\left(y-z\right)\)
\(=\left(y-z\right)\left(xy+zx-yz-x^2\right)\)
\(=\left(y-z\right)\left[\left(zx-yz\right)-\left(x^2-xy\right)\right]\)
\(=\left(y-z\right)\left[z\left(x-y\right)-x\left(x-y\right)\right]\)
\(=\left(x-y\right)\left(y-z\right)\left(z-x\right)\)
x(y+z)^2 - y(z-x)^2 +z(x+y)^2 - x^3 + y^3 - z^3 - 4xyz
=xy^2+2xyz+xz^2-yz^2+2xyz-x^2y+x^2z+2xyz+zy^2-x^3+y^3-z^3-4xyz
=xy^2+xz^2-yz^2-x^2y+x^2z+y^2z-x^3+y^3-z^3+2xyz
=(xy^2+2xyz+xz^2)-x^3-(yz^2+2xyz+x^2y)+y^3+(x^2z+2xyz+y^2z)-z^3
=x[(y+z)^2-x^2)-y[(z+x)^2-y^2]+z[(x+y)^2-z^2]
=x(-x+y+z)(x+y+z)-y(x-y+z)(x+y+z)+z(x+y-z)(x+y+z)
=(x+y+z)[-x^2+xy+xz-xy+y^2-yz+xz+yz-z^2]
=(x+y+z)[-x(x-y-z)-y(x-y-z)+z(x-y-z)]
=(x+y+z)(x-y-z)(z-x-y)
Câu hỏi của nguyễn khánh linh - Toán lớp 8 - Học toán với OnlineMath
a,Từ giả thiết ta có
(x2+y2+z2)(x+y+z)2+(xy+yz+zx)2
=(x2+y2+z2)(x2+y2+z2+2xy+2yz+2zx)+(xy+yz+zx)2
Đặt x2+y2+z2=a
xy+yz+zx=b
=>(x2+y2+z2)(x2+y2+z2+2xy+2yz+2zx)+(xy+yz+zx)2
=a(a+2b)+b2
=a2+2ab+b2
=(a+b)2
=(x2+y2+z2+xy+yz+zx)2
câu b hơi dài mình gửi sau nhé
Ta có: 2(x^4+y^4+z^4)-(x^2+y^2+z^2)^2-2(x^2+y^2+z^2)(x+y+z)^2+(x+y+z)^4
Gọi x^4+y^4+z^4=a
x^2+y^2+z^2=b
x+y+z=c
=>2(x^4+y^4+z^4)-(x^2+y^2+z^2)^2-2(x^2+y^2+z^2)(x+y+z)^2+(x+y+z)^4=2a-b^2-2bc^2+c^4
=2a-2b^2+b^2-2bc^2+c^4
=2(a-b^2)+(b+c^2)^2
Ta có
2(a-b2)=2[x^4+y^4+z^4-(x^2+y^2+z^2)2]
=2[x^4+y^4+z^4-x^4-y^4-z^4-2x2y2-2y2z2-2z2x2]
=2.(-2)(x2y2+y2z2+z2x2)
=-4(x2y2+y2z2+z2x2)
Lại có
(b+c^2)^2
=[(x^2+y^2+z^2)+(x+y+z)2]2
=[(x^2+y^2+z^2)-(x^2+y^2+z^2)-2(xy+yz+zx)]2
=4(xy+yz+zx)2
=>2(a-b^2)+(b+c^2)^2
=-4(x2y2+y2z2+z2x2)+4(xy+yz+zx)2
=8xyz(x+y+z)
x(y+z)^2 - y(z-x)^2 +z(x+y)^2 - x^3 + y^3 - z^3 - 4xyz
=xy^2+2xyz+xz^2-yz^2+2xyz-x^2y+x^2z+2xyz+zy^2-x^3+y^3-z^3-4xyz
=xy^2+xz^2-yz^2-x^2y+x^2z+y^2z-x^3+y^3-z^3+2xyz
=(xy^2+2xyz+xz^2)-x^3-(yz^2+2xyz+x^2y)+y^3+(x^2z+2xyz+y^2z)-z^3
=x[(y+z)^2-x^2)-y[(z+x)^2-y^2]+z[(x+y)^2-z^2]
=x(-x+y+z)(x+y+z)-y(x-y+z)(x+y+z)+z(x+y-z)(x+y+z)
=(x+y+z)[-x^2+xy+xz-xy+y^2-yz+xz+yz-z^2]
=(x+y+z)[-x(x-y-z)-y(x-y-z)+z(x-y-z)]
=(x+y+z)(x-y-z)(z-x-y)